If secθ = 29⁄20 where 0 < θ < 90°, then what is the value of 3cosecθ + 3cotθ?

Correct — A. sec θ = 29⁄20 fixes the triangle at once: hypotenuse 29, adjacent 20.

Third side = √(29² − 20²) = √(841 − 400) = √441 = 21
The range 0 < θ < 90° keeps every ratio positive, so the side is +21.

cosec θ = 29⁄21 and cot θ = 20⁄21

3 cosec θ + 3 cot θ = 3 × (29⁄21 + 20⁄21)
= 3 × 49⁄21
= 147⁄21 = 7 → option (a)

Why the others are wrong

• (b)
  49 — 49 is 29 + 20 with the division never done. The common denominator 21 is still owed: 3 × 49 ⁄ 21 = 7, not 49.
• (c)
  14 — 14 is twice the correct value. The division is by 21, the side opposite θ, and 147 ⁄ 21 is 7 — nothing in the working halves that denominator.
• (d)
  7/3 — 7⁄3 is cosec θ + cot θ on its own, since 49⁄21 = 7⁄3. The stem multiplies the whole sum by 3, and 3 × 7⁄3 = 7.

One ratio pins down the whole right triangle. sec θ names the hypotenuse over the adjacent side, so 29⁄20 hands you two sides and Pythagoras supplies the third.

20, 21, 29 is a Pythagorean triple worth memorising, because recognising it saves the square roots: 400 + 441 = 841.

The condition 0 < θ < 90° is not decoration. It puts θ in the first quadrant, where all six ratios are positive, so the third side is taken as +21 and no sign case has to be considered.

Key facts

• sec θ is hypotenuse ⁄ adjacent, so sec θ = 29⁄20 means hypotenuse 29 and adjacent side 20.
• 20, 21, 29 is a Pythagorean triple: 400 + 441 = 841.
• For 0 < θ < 90° every trigonometric ratio is positive, so the third side is +21.
• cosec θ + cot θ = (1 + cos θ) ⁄ sin θ, which here evaluates to 7⁄3.

Common traps

• Reading sec θ = 29⁄20 as cos θ = 29⁄20 and building the triangle inverted.
• Subtracting the sides, 29 − 20 = 9, instead of taking √(29² − 20²) = 21.
• Stopping at 7⁄3 and forgetting the factor 3 that multiplies both terms.

No directly related past PYQ was found.